(q-1)(q-5)=-3

Solution for (q-1)(q-5)=-3 equation:

(q-1)(q-5)=-3

We move all terms to the left:

(q-1)(q-5)-(-3)=0

We add all the numbers together, and all the variables

(q-1)(q-5)+3=0

We multiply parentheses ..

(+q^2-5q-1q+5)+3=0

We get rid of parentheses

q^2-5q-1q+5+3=0

We add all the numbers together, and all the variables

q^2-6q+8=0

a = 1; b = -6; c = +8;
Δ = b2-4ac
Δ = -62-4·1·8
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2}{2*1}=\frac{4}{2}
=2
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2}{2*1}=\frac{8}{2}
=4
$